# A student is throwing rocks off of a bridge straight down into a river below.?

A student is throwing rocks off of a bridge straight down into a river below. If he throws a rock with an initial speed of 10 m/s and it takes 2.1s to hit the water, how high is the bridge? Ignore Air resistance.

Relevance

Distance = -g/2×(t)^2 - v×(t)

Distance = - 4.9× (2.1)^2 - 10×(2.1) = -42.6 m

minus means below bridge.

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• y(t) = ½at² +v₀t + h₀. <<<< memorize this. it's used extensively!!!

y is the distance

a is the acceleration, gravity in this case

t is time and must agree with the time units of the other factors

v₀ is the initial velocity, 10m/s

h₀ is initial height or position, in this case zero

'a' s Earth Standard Gravity in most projectile flights (but not always!!!)

The precise strength of Earth's gravity varies depending on location. The nominal "average" value at the Earth's surface, known as "standard gravity" is, by definition, 9.80665 m/s² (about 32.1740 ft/s²).

y(t) = ½at² +v₀t + h₀ put in your known values

0 = ½ * -9.80665 * (2.1)² -10*2.1 + h

notice 'down' is a negative number

h = 21.62366325 + 21

h = 42.62366325 m

Sigfigs: 2

h = 43 m

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• THROWING or dropping---there is a difference---Throwing would imply extra velocity was being added to the rocks.

• The question is not trying to imply extra velocity

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• Your student should have read the sign "No throwing rocks off this bridge. Penalty includes a fine of \$1000 and 1 month in prison".

s = ut + 1/2 a t^2

= -10 * 2.1 - 1/2 * 9.8 * 2.1^2 ~= - 43m

ie it FELL 43 m so the bridge was 43 m high.

Note that as the time was only quoted as 2.1 s the distance is actually "between 41.6 and 43.6 m"

2.1 is not identical to 2.10 or 2.100

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• Use the equation of motion:

y(t)= y0 + v0t + 1/2 g t^2

y(t) is the height at any time t, v0 is the initial velocity, g is the accel due to gravity, and t is the time

for a problem like this, be careful how you choose signs; let's say down is the positive direction (we can make any choice for positive we wish, but we have to be careful to be consistent)

let's set y = 0 to be the launch point, then going down is increasing in distance,

thus, we want to know what is the value of y when t = 2.1 s

we have:

y(t = 2.1s) = 0+ 10 m/s * 2.1 s + 1/2 g (2.1s)^2

solve this equation and get the distance from the launch point to the river

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• Vf = Vo+g*t = 10+9.806*2.1 = 30.593 m/sec

conservation of energy shall apply :

Vf^2 = Vo^2+2gh

h = (Vf^2-Vo^2)/2g = (935.91-100)/19.612 = 42.62 m

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• Let’s use the following equation to calculate the distance the ball falls in 2.1 seconds.

d = vi * t + ½ * a * t^2

The ball has a vertical acceleration of -9.8 m/s^2.

d = 10 * 2.1 + ½ * -9.8 * 2.1^2 = -0.609 meter

This is the vertical displacement of the ball. The negative sign means the ball final height 0.609 meter below its initial height. I hope this is helpful for you. So the height of the bridge is 0.609 meter.

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